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Ex:  $P^n(\R) = \R P^n$ =   $\R P^n =( \R^n - \{\0\}  )/ (\x \sim t\x)$ \hfil \break = $n$-dimensional real projective space is a smooth manifold.
 


{\bf Claim:  $\R P^n$ is 2nd countable.}

We will use \hfil \break
Lemma: If $\sim$ is {open} and if $X$ has a countable basis, then $X/\sim$  has a countable basis.

[{We will define a map which takes $\y \in [\x]$ to $t\y \in [\x]$ }]

Let $\phi_t:   \R^n - \{\0\}   \rightarrow \R^n - \{\0\}  $, 
$\phi_t(\x) = t\x$. 

$\phi_t$ is invertible with inverse $\phi_t^{-1} =  \phi_{1 \over t}$.

Since $\phi_t$ and $\phi_t^{-1}$ are $C^1$ (as well as $C^{\infty}$), $\phi_t$ is a homeomorphism.

Let $U$ be open in  $\R^n - \{\0\} $.  Then $\phi_t(U)$ is open in $\R^n - \{\0\} $.

Thus $\pi^{-1}([U]) = \cup_{t \in \R} \phi_t(U)$ is open in $\R^n - \{\0\} $.  

Thus $[U]$ is open in $\R P^n$.  Hence $\sim$ is open.  

Since $\R^n$ is 2nd countable, $\R P^n$ is 2nd countable.


{\bf Claim $\R P^n$ is Hausdorff}

We will use 
\hfil \break
Lemma:  Let $\sim$ be {open}.  Then $\{ (x, y) ~|~ x \sim y \}$ is closed in $X \times X$ iff $X/\sim$  is Hausdorff.

[We will show that $\{ (x, y) ~|~ x \sim y \} = f^{-1}(\{0\})$ for some continuous function $f$.  $x \sim y$ implies $x_i = ty_i$.  Thus ${x_i \over y_i} = {x_j \over y_j}$.  Hence $x_iy_j - y_ix_j = 0$ for all $i, j$.]

Let $f:  \R^n - \{\0\}  \times  \R^n - \{\0\}  \rightarrow \R$, \hfil \break
$f(x_1, ..., x_n, y_1, ..., y_n) = \Sigma_{i \not= j} (x_iy_j - y_ix_j )^2$.

$f$ is $C^1$ (all partials of $f$ exist and are continuous).  Thus $f$ is continuous.

Suppose $\y = t\x$, then  \hfil \break
$f(x_1, ..., x_n, y_1, ..., y_n) = \Sigma_{i \not= j} (x_itx_j - tx_ix_j )^2 = 0$.

Suppose $f(x_1, ..., x_n, y_1, ..., y_n) = \Sigma_{i \not= j} (x_iy_j - y_ix_j )^2 = 0$.
Then $x_iy_j - y_ix_j = 0$ for all $i, j$.  Since $\x \not= \0$, there exists $i_0$ such that $x_{i_0} \not= 0$.  Thus $y_j = {y_{i_0} \over x_{i_0}} x_j$ and $\y = {y_{i_0} \over x_{i_0}} \x$.  Hence $\x \sim \y$.

Hence  $f^{-1}(\{0\}) = \{ (x, y) ~|~ x \sim y \} $.  Since $f$ is continuous and $\{0\}$ is closed in $\R$,  $\{ (x, y) ~|~ x \sim y \}$ is closed in $ \R^n - \{\0\}  \times  \R^n - \{\0\} $.  Thus $ \R^n - \{\0\} /\sim$    is Hausdorff.

\vskip 5pt
\hrule

We will show that $\R P^n$ is locally Euclidean by finding a (pre) atlas:

Let $V_i = \{x \in \R^n - \{\0\} ~|~ x_i \not= 0 \} \subset \R^n - \{\0\}$ 

Let $F_i: V_i \rightarrow \R^n$, 
$F_i(x_1, ..., x_{n+1}) = ( {x_1 \over x_i}, 
{x_2 \over x_i}, ..., {x_{i-1} \over x_i}, {x_{i+1} \over x_i}, ..., {x_{n+1} \over x_i})$


\rightline{$= {1 \over x_i}(x_1, ..., \hat{x_i}, ..., x_{n+1})$} 


$F_i(t\x) = ( {tx_1 \over tx_i}, 
{tx_2 \over tx_i}, ..., {tx_{i-1} \over tx_i}, {tx_{i+1} \over tx_i}, ..., {tx_{n+1} \over tx_i})$

\rightline{$= ( {x_1 \over x_i}, 
{x_2 \over x_i}, ..., {x_{i-1} \over x_i}, {x_{i+1} \over x_i}, ..., {x_{n+1} \over x_i})$
$= F_i(\x)$. }


Let $U_i = \pi(V_i)$
Then $\phi_i:  U_i \rightarrow \R^n$, $\phi_i([\x]) = F_i(\x)$ is well-defined.

\eject

{\bf Claim:  $(\phi_i, U_i)$ is a chart.}



~~~~~~~~~~~~{\bf Subclaim 1:  $U_i$ is open in $\R P^n$.}

$\pi_i^{-1}(U_i) = \pi_i^{-1}(\pi(V_i)) = V_i$ [by set theory].  Hence  $U_i$ is open in $\R P^n$.

~~~~~~~~~~~~{\bf Subclaim 2:  $\phi(U_i)$ is open in $\R^n$.}

Claim:  $\phi_i$ is onto.  

Let $(x_1, ..., x_n) \in \R^n$.  
$\phi_i([(x_1, ..., x_{i-1}, 1, x_i, .., x_n]) = (x_1, ..., x_n)$.  Thus $\phi_i$ is onto.

Since $\phi(U_i) = \R^n$ , $\phi(U_i)$ is open in $\R^n$.


~~~~~~~~~~~~{\bf Subclaim 3:  $\phi_i$ is a homeomorphism.}

Claim:  $\phi_i$ is continuous.

Let $V$ be open in $\R^n$.  \hfil \break
Since $F_i \in C^1$,  $F_i^{-1}(V)$ is open in $\R^{n+1} - \{\0\}$.

$\pi^{-1}\circ \phi_i^{-1}(V) = F_i^{-1}(V)$.  Thus $\phi_i^{-1}(V) $ is open in $U_i$ and 
$\phi_i$ is continuous.


Claim:  $\phi_i$ is 1:1.


If $\phi_i([\x])$  = $\phi_i([\y])$, then ${x_j \over x_i} = {y_j \over y_i}$ for all $j$.  Thus $y_i =  {y_i \over x_i}x_j$ and thus $\y = {y_i \over x_i}\x$.  Thus  $\phi_i$ is 1:1.

Since $\phi_i$ is 1:1 and onto, $\phi_i^{-1}$ exists.

\eject

Claim:  $\phi_i^{-1}: \R^n \rightarrow \R P^n$ is continuous.

 $\phi_i^{-1}(x_1, ..., x_n) = [(x_1, ..., x_{i-1}, 1, x_i, .., x_n)]$. 



Let $f_i: \R^n \rightarrow \R^{n+1}$, $f_i(\x) = (x_1, ..., x_{i-1}, 1, x_{i}, ..., x_n)$.  
$f_i$ is $C^1$ and hence continuous.  

$\pi \circ f(\x) = [(x_1, ..., x_{i-1}, 1, x_i, .., x_n)] = \phi_i^{-1}(x_1, ..., x_n)$. 

Since $\pi$ and $f$ are continuous,  $\phi_i^{-1}: \R^n \rightarrow \R P^n$ is continuous.

Thus $(\phi_i, U_i)$ is a chart.

{\bf Claim:  $\{(\phi_i, U_i) ~|~ i = 1, ..., n+1\}$ is a (pre) atlas for $\R P^n$.}

$\R^{n +1} - \{\0\} = \cup_{i=1}^{n+1} V_i$.  Thus $\R P^n = \cup_{i=1}^{n+1} U_i$.  


{\bf Claim:  $\phi_j \phi_i^{-1}:  \phi_i(U_i \cap U_j) \rightarrow \phi_i(U_i \cap U_j)$ is smooth.}

Suppose $j < i$.

$\phi_j(\phi_i^{-1}(x_1, ..., x_n)) = \phi_j([x_1, ..., x_{i-1}, 1, x_{i}, ..., x_n])$

\rightline{$ = 
({x_1 \over x_j} , ..., {x_{i-1} \over x_j} , {1 \over x_j} , {x_{i} \over x_j} , ..., {x_n \over x_j} )$}



Since all the components of $\phi_j \phi_i^{-1}$ are polynomials, $\phi_j \phi_i^{-1}$ is smooth.

Similarly $\phi_j \phi_i^{-1}$ is smooth when $j > i$.


Thus $\{(\phi_i, U_i) ~|~ i = 1, ..., n+1\}$ is a (pre) atlas for $\R P^n$,
and $\R P^n$ is a smooth manifold.




   



\end

${\partial \over x_k}  ({x_{m} \over x_j}) = \cases{0 & $k \not= m, j$ \cr 
{1 \over x_j} & $ k = m$ \cr {-x_{m} \over x_j^2} & $m = j$}$
\hfill
${\partial \over x_k}  ({1 \over x_j}) = \cases{0 & $k \not= j$ \cr 
 {-1 \over x_j^2} & $m = j$}$

Since all partials exist and are continuous, $\phi_j(\phi_i^{-1}$ is C^1